Buns（dp+多重背包）

最新推荐文章于 2021-08-06 16:22:18 发布

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最新推荐文章于 2021-08-06 16:22:18 发布

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原文链接：http://www.cnblogs.com/handsomecui/p/5255180.html

本文介绍了一个烘焙问题的解决方法，通过使用多重背包算法确定在有限的面团和馅料条件下，制作不同类型的糕点以获得最大收益。文章详细阐述了算法实现过程。

Lavrenty, a baker, is going to make several buns with stuffings and sell them.

Lavrenty has n grams of dough as well as m different stuffing types. The stuffing types are numerated from 1 tom. Lavrenty knows that he has a_i grams left of the i-th stuffing. It takes exactly b_i grams of stuffing i and c_igrams of dough to cook a bun with the i-th stuffing. Such bun can be sold for d_i tugriks.

Also he can make buns without stuffings. Each of such buns requires c₀ grams of dough and it can be sold ford₀ tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.

Find the maximum number of tugriks Lavrenty can earn.

Input

The first line contains 4 integers n, m, c₀ and d₀ (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10, 1 ≤ c₀, d₀ ≤ 100). Each of the following m lines contains 4 integers. The i-th line contains numbers a_i, b_i, c_i and d_i (1 ≤ a_i, b_i, c_i, d_i ≤ 100).

Output

Print the only number — the maximum number of tugriks Lavrenty can earn.

Examples

input

10 2 2 1 7 3 2 100 12 3 1 10

output

input

100 1 25 50 15 5 20 10

output

Note

To get the maximum number of tugriks in the first sample, you need to cook 2 buns with stuffing 1, 4 buns with stuffing 2 and a bun without any stuffing.

In the second sample Lavrenty should cook 4 buns without stuffings.

思维：让你做蛋糕，有各种口味的奶油，和面，消耗一定的面一定口味的奶油可以卖d元，问最多买多少钱；因为是蛋糕肯定是整个整个的，所以要用到背包；多重背包；

dp[i][k]代表前i种蛋糕在面小于等于k的最大利润；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
struct Node{
    int a,b,c,d;
};
Node dt[15];
int dp[15][1010];
int main(){
    int n;
    int m;
    while(~scanf("%d%d%d%d",&n,&m,&dt[0].c,&dt[0].d)){
        dt[0].a=100010;dt[0].b=1;
        for(int i=1;i<=m;i++)scanf("%d%d%d%d",&dt[i].a,&dt[i].b,&dt[i].c,&dt[i].d);
        mem(dp,0);
        for(int i=0;i<=m;i++){
            for(int j=0;j*dt[i].b<=dt[i].a;j++){
                for(int k=n;k>=j*dt[i].c;k--){
                    if(i)dp[i][k]=max(dp[i][k],dp[i-1][k-j*dt[i].c]+j*dt[i].d);
                    else dp[i][k]=max(dp[i][k],dp[i][k-j*dt[i].c]+j*dt[i].d);
                }    
            }
        }
        printf("%d\n",dp[m][n]);
    }
    return 0;
}